Lemma 21.17.16. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{K}^\bullet $ be a K-flat, acyclic complex with flat terms. Then $\mathcal{F} = \mathop{\mathrm{Ker}}(\mathcal{K}^ n \to \mathcal{K}^{n + 1})$ is a flat $\mathcal{O}$-module.
Proof. Observe that
\[ \ldots \to \mathcal{K}^{n - 2} \to \mathcal{K}^{n - 1} \to \mathcal{F} \to 0 \]
is a flat resolution of our module $\mathcal{F}$. Since a bounded above complex of flat modules is K-flat (Lemma 21.17.8) we may use this resolution to compute $\text{Tor}_ i(\mathcal{F}, \mathcal{G})$ for any $\mathcal{O}$-module $\mathcal{G}$. On the one hand $\mathcal{K}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{G}$ is zero in $D(\mathcal{O})$ because $\mathcal{K}^\bullet $ is acyclic and on the other hand it is represented by $\mathcal{K}^\bullet \otimes _\mathcal {O} \mathcal{G}$. Hence we see that
\[ \mathcal{K}^{n - 3} \otimes _\mathcal {O} \mathcal{G} \to \mathcal{K}^{n - 2} \otimes _\mathcal {O} \mathcal{G} \to \mathcal{K}^{n - 1} \otimes _\mathcal {O} \mathcal{G} \]
is exact. Thus $\text{Tor}_1(\mathcal{F}, \mathcal{G}) = 0$ and we conclude by Lemma 21.17.15. $\square$
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